What is the force F on the 8.0 nC charge in FIGURE P22.44? Give y...

Question

What is the force F on the 8.0 nC charge in FIGURE P22.44? Give your answer as a magnitude and an angle measured cw or ccw (specify which) from the +x-axis.

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Hello, everyone. Let's take a look at this practice problem. We have three small balloons with static electric charges in a square arrangement. The 1st 2nd and 3rd balloons have charges of negative eight nano cums, positive six nano columns and positive eight nano columns respectively. Compute the magnitude and angle clockwise or counterclockwise from the positive X axis of the net force acting on the six nano Coolum charged balloon. And we have four multiple choice options to choose from option A 3.1 multiplied by 10 to the power of a negative three newtons. 53 degrees clockwise from the positive X axis. Option B 7.3 multiplied by 10 to the power of negative four newtons. 58 degrees clockwise. Option C 2.8 multiplied by 10 to the power of negative three new ones, 33 degrees or option D 4.8 multiplied by to the power of negative four new ones 23 degrees because we're looking for the net force acting on the six and nano Coolum charged balloon. So in other words, Q sub two, we need to figure out how the electrostatic forces from the two other charges, Q one and Q three are going to impact it. So let's consider what we know about the, about the law of attraction between charged objects. We know that like objects repel each other and opposites attract. So if we look at the Q, the, this, the balloon, we're looking at Q sub two, we can see that Q sub one is negatively charged while Q sub two is positively charged, that means that there is going to be an attractive force between these two charges. So Q sub two is going to experience a downwards force due to the attraction between Q one and Q two. We're just going to use a standard coordinate system for this. There's also, it's also going to experience an electrostatic force from Q sub three, which is also positively charged. That means that there is going to be a repulsive force between Q three and Q two. So Q two is going to experience a repulsive force directed away diagonally from Q sub three. Let's now analyze these forces using Kums Law and recall that Kums Law gives us a relationship between the charges and the force between them. So according to Cole's law, the force between two electrically charged object is equal to the Coolum constant multiplied by the product of the two charges we're looking at divided by the square of the distance between them for the purpose of this problem. Let's start by looking at the force on charge two from charge one. So let's start with that. So we're given the magnitudes of all the charges. So that part is going to be fairly simple since we're given charge one as eight nano columns and charge two as six nano columns before I start filling things in, I want to note real quick that because this is a more complex system, it can be a little confusing to take the signs into consideration super early. It's important to recognize which direction the the particle will be tended to accelerate due to each charge. When we're setting up force equations like this, it's sometimes helpful to only focus on the magnitudes at first. So I'm not going to be taking into account of the negatives on the charges for the purposes of these Kulum law expressions. But the magnitude from charge one on charge two is equal to the Coolum constant, which has a value of 8.99 multiplied by 10 to the power of nine Newton meters squared per Coolum squared multiplied by the product of the charges. So Q sub one or 8. nano Kums alternatively eight multiplied by 10 to the power of a negative nine columns multiplied by Q sub nano columns or six multiplied by 10 to the power of negative nine columns. This is all divided by the square of the distance between them. So the square of two centimeters or 0.02 m. And if you put that into a calculator, we find a force of about negative 1. multiplied by 10, the power of negative three nuance, the negative sign in front of this value could come from the fact that Q one is negatively charged. But we know it has to be negative anyway, because of the reason we discussed earlier that we know Q two is going to be tended to be pulled downwards towards the negative charge. This means that since this is entirely along the Y axis, this is along the J unit vector. Now let's do this exact same thing before the force on charge two due to the presence of charge three. So I'm gonna call it F sub three comma two. This is going to be the force of the Coolum force multiplied by the product of Q sub three and Q sub two divided by the distance between particles three and two squared. We're not given that distance. But we can see from the diagram that it's essentially going to be the hypotenuse of a triangle formed by the sides of this square arrangement. So that means that R sub 32 is equal to the square root of the sum of the squared of the two signs. So two centimeters squared plus two centimeters squared or 0. m squared plus 0.02 m squared. Now let's put this into our formula. So F sub 32 as a value of the Coolum constant 8.99 multiplied by 10 of the power of nine new and meter squared per colum squared multiplied by the product of the charges or eight multiplied by 10 to the power of negative nine columns, multiplied by six, multiplied by 10 of the power of negative nine columns, all divided by the square of R sub 32, which basically just cancels out the square root. So 0.02 m squared plus 0.02 m squared. And if we put this into a calculator, we find a force magnitude of 5. multiplied by 10 to the power of negative four newtons. Now, we can't apply a uh a unit vector to this answer because it's directed diagonally. But if we're gonna want to find the net force acting on the particle, we're gonna want to break this up into components. So we first need to find F sub 32 X, the X component of that force as well as F sub Y, the Y component of that force. And we can find this by simply taking either our taking our magnitude of the force and then multiply it by the sign or the cosine of the angle within this square. And it's convenient for us because since it's a square shape, that means that this angle must be 45 degrees. So we don't even need to do any trigonometry to figure that out. But this X component, for example, is going to be F sub 32 multiplied by the cosine of 45 degrees. So we can put this into a calculator as 5.4 multiplied by of the power of negative four new ones, all multiplied by the cosine of 45 degrees. Put this into a calculator. We find a value of 3.81 multiplied by 10 to the power of a negative four nuances. And this must be positive in the I hat direction because of the fact that they're both positive. So part particle two is gonna be propelled towards the right in the positive X direction, I'll apply the same logic to the Y component. So 5.4 multiplied by 10 to the power of negative four nuances multiplied by the sine of 45 degrees. Put this into a calculator. And we get the exact same magnitude which is natural because the sign of co of 45 degrees is the exact same as the cosine of 45 degrees only. Now, since we're looking at the Y component, it is directed positive with AJ hat unit vector. So now that we have all of our forces in terms of components, now let's add these components together to find the net components acting on particle two. So for F sub 32, I'm sorry, F sub two X. So the net horizontal force is acting on particle two, we only have, we just have this 3.81 here because the other force components we found had the J hat unit vector. So that part remains unchanged. 3.81 multiplied by 10 to the power of negative four nus in the eye direction. As for F sub two Y, we're going to have to add together the J hat we just found for 32 as well as the J hat we found for 12. So from one, it was a negative one multiplied by 10 to the power of negative three nus, let's write that out negative one multiplied by 10 to the power of negative three nubs plus than the one we found from particle 33.81 multiplied by 10 of the power of negative four new MS and adding this together the net force in the Y direction is negative 6.1 F 19 multiplied by 10 to the power of negative four new ones all in EJ direction. So lastly, if we want to find the magnitude of the force acting on particle two, we simply have to take the, use the Pythagorean theorem method take the square root of the sum of the squares of both components. So the square root of 3. multiplied by 10 to the power of negative four newtons, all squared plus the square of negative 6.19 multiplied by 10 to the power of negative four newtons. And if we put this into a calculator. We find a net force magnitude of about 7. multiplied by 10, raise the power of negative four nuances. So that is our magnitude. But we still need to find the angle from the positive X axis. So recall that the formula for this is to take the negative the inverse tangent of the Y component divided by the X component. That's the inverse tangent of negative 6.19 multiplied by 10 to the power of negative four new ones, all divided by 3.81 multiplied by 10 to the power of the negative four newtons. If we put this into a calculator, then we find an angle here of approximately negative degrees. And if it's negative, that means that the angle is measured clockwise from the positive X axis. So we have now officially found both the magnitude and the angle of our net vector. And if we look at our multiple choice options, we can see that option B agrees with what we found 7. multiplied by 10 to the power of negative four newtons, 58 degrees clockwise from the positive X axis. So that is our answer to this problem. Option B is the answer and that is it for this problem. I hope this video helped you out and please consider checking out some of our other videos which will give you more experience with these types of problems. But that's all for now and I hope you all have a lovely day. Bye bye.

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What is the force F on the 8.0 nC charge in FIGURE P22.44? Give your answer as a magnitude and an angle measured cw or ccw (specify which) from the +x-axis.

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